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Work and kinetic energy

  • #1
A rocket car is travelling at 648km/h[f] when the parachute is deployed and does 51.64MJ of work to slow the car down to a speed of 54.o km/h



Homework Equations



w = delta ek
w= m(1/2mvf^2 - 1/2mvi^2)
m= w/0.5(Vf^2-Vi^2)

The Attempt at a Solution



i derived that equation from w=delta kinetic energy and i got 2/3 for showing work... this was a test question 4 months ago

now i have an exam tomorrow.

the mass i got is 248kg. the teacher took it up and the correct mass was 3210kg. i honestly feel my answer is correct and i constantly repeat this question and get 248 kg!!!

he said my formula is correct just the calculation is wrong:grumpy:

this is what i did.

51.64x10^6J/ -208494(negative since Vi > Vf so you get negative mass but i guess you ignore the negative)

can anyone tell me what i am doing wrong?!?!???!
 

Answers and Replies

  • #2
gneill
Mentor
20,793
2,773
Recheck your value for v02 - vf2.
 
  • #3
verty
Homework Helper
2,164
198
If the velocities were 648 m/s and 54 m/s, the answer would be 248 kg.

Let 1 km/h = c m/s

[tex]m = 2w / (v_1^2 - v_2^2)[/tex]
[tex] = 2 * 51.64 * 10^6 / (648^2 c^2 - 54^2 c^2)[/tex]
[tex] ~= 248 / c^2[/tex]

This formula only works for velocities in m/s, we see.
 
  • #4
If the velocities were 648 m/s and 54 m/s, the answer would be 248 kg.

Let 1 km/h = c m/s

[tex]m = 2w / (v_1^2 - v_2^2)[/tex]
[tex] = 2 * 51.64 * 10^6 / (648^2 c^2 - 54^2 c^2)[/tex]
[tex] ~= 248 / c^2[/tex]

This formula only works for velocities in m/s, we see.

i am no expert in physics but i think you made mistake. shouldn't it be 54^2-648^2 not the way you wrote it since it is slowing down. you subtracted v initial from v final... the formula is v final minus vinitial
 
  • #5
verty
Homework Helper
2,164
198
i am no expert in physics but i think you made mistake. shouldn't it be 54^2-648^2 not the way you wrote it since it is slowing down. you subtracted v initial from v final... the formula is v final minus vinitial
[tex]w = \Delta E_k = (1/2) m (v_f^2 - v_i^2)[/tex]

I should have made the work negative because the change of energy was negative, energy was lost. So I made two changes to the formula, I neglected to make the work value negative, but I also wrote the velocities back to front. I could do that for this formula because it made no difference.

You can see that because:

[tex]\frac{2w}{v_f^2 - v_i^2} = \frac{2(-w)}{v_i^2 - v_f^2}.[/tex]

So I actually used the second part of that, because I used -w (by using 51MJ and not -51MJ). Admittedly, I should have used -51 MJ and done it the correct way round, just to be safe.

I only did it because I could do it without making the answer invalid. But as you saw, what I could not do was use km/h values for the velocities.
 

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